Question

A stone of mass $0.25kg$ tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40rev./min$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200N$?

Answer 1

Mass of the stone, $m=0.25kg$

Radius of the circle, $r=1.5m$
Number of revolution per second, $n=40/60=2/3rps$
Angular velocity, $\omega =2\pi n$

The centripetal force for the stone is provided by the tension $T$, in the string, i.e.,

$T=m{\omega }^{2}r$
$=0.25\times 1.5\times (2\times 3.14\times (2/3){)}^2$
$=6.57N$
Maximum tension in the string, $T_{max}=200N$
$T_{m}ax=\frac{m{v}_{max}^2}{r}$
$v_{max}={\left(\frac{T_{max}r}{m}\right)}^{1/2}$

$=(200\times 1.5/0.25{)}^{1/2}$

$=\left(1200\right)$${}^{1/2}$ = $34.64m/s$

Therefore, the maximum speed of the stone is $34.64m/s$.